K_1, K_2 and K_3 are the equilibrium constant | vedclass

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(B) For reaction $(I): N_2 + 2O_2 \rightleftharpoons 2NO_2$,the equilibrium constant is $K_1 = \frac{[NO_2]^2}{[N_2][O_2]^2} \dots (i)$For reaction $(

Vedclass · Accepted Answer

$K_1 = \frac{1}{K_2} = \frac{1}{(K_3)^2}$

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(B) For reaction $(I): N_2 + 2O_2 \rightleftharpoons 2NO_2$,the equilibrium constant is $K_1 = \frac{[NO_2]^2}{[N_2][O_2]^2} \dots (i)$For reaction $(II): 2NO_2 \rightleftharpoons N_2 + 2O_2$,which is the reverse of $(I)$,the equilibrium constant is $K_2 = \frac{[N_2][O_2]^2}{[NO_2]^2} = \frac{1}{K_1} \dots (ii)$For reaction $(III): NO_2 \rightleftharpoons \frac{1}{2} N_2 + O_2$,the equilibrium constant is $K_3 = \frac{[N_2]^{1/2}[O_2]}{[NO_2]}$Squaring $K_3$,we get $(K_3)^2 = \frac{[N_2][O_2]^2}{[NO_2]^2} = K_2 = \frac{1}{K_1} \dots (iii)$Thus,$K_1 = \frac{1}{K_2} = \frac{1}{(K_3)^2}$.

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